http://books.google.com/books?id=x1q...hat%22&f=false
72a. I have no frigging clue where to start.
http://books.google.com/books?id=x1q...hat%22&f=false
72a. I have no frigging clue where to start.
isn't this a Riemann summation? It's been a longtime since I've done that, but it looks really similar.
Just bumping this post up to say no. @Locked, you have any ideas?
Where's @Noah when you need him?
Other than it being a seemingly useless proof, not really.
My friend proved it though. I'll tell him to post here.
By useless, I mean that it doesn't make anything easier really. You could do the second problem (72b) without proving anything and just using regular partial fractions. Though that ruins the spirit of the problem ;p
Full proof:
Additional resource:
Basically just do the same thing you do for a normal partial fraction problem, just with 10x more symbols and a little bit of reasoning. There's a sketchy part in expressing the product from m=1 to j1 when j can very well be 1. I can easily do a piecewise solution (when j = 1, when j = n, and otherwise), but you get the idea of the proof anyway. And by "two of the three terms", I mean either the A1 term remains, or the An term remains, or Aj remains where 1 < j < n. Better wording should say "all but one term get eliminated" instead.
EDIT: Long time no see, guys.
Last edited by Kalovale; 20121026 at 11:06 PM.
Here's an English translation:
 Step 1: Clear the denominator by (1) multiplying each numerator (A1, A2, ..., Aj, ..., An) with every (x  aj) except for the one directly below it and (2) summing them together.
 Step 2: Notice that each term of this pattern: Aj * (x  a[everything else except j]), thus the only term that is left when plugging in x = aj is the one that does not have (x  aj) in it. Incidentally, this is the Aj term.
 Step 3: So P(aj) = Aj * (aj  a1) * (aj  a2) * ... * (aj  aj1) * (aj  aj+1) * ... * (aj  an)
 Step 4: Take the derivative of Q(x), using product rule after reordering Q(x) into (x  aj) * everything else. Q'(x) = (x  aj)' * everything else + (x  aj) * (everything else)'
 Step 5: The second part is 0 when plugging in x = aj, so Q'(aj) = everything else = (aj  a1) * (aj  a2) * ... * (aj  aj1) * (aj  aj+1) * ... * (aj  an)
 Step 6: P(aj)/Q'(aj) = Aj
I can't see the page for some reason, but if you were trying to isolate Q'(x) I think the 2nd method is a better proof, academically speaking. Also in terms of clarity.
Hadriel
Ah ok I see why that was necessary now.
Hadriel
I feel like we need more math questions in here.
In this thread?
I don't have that fancy mathwriting program, sorry if it's sloppy :(
f(x,y) = xy + 64/x + 64/yOriginally Posted by Question
Local Maximum values(s) : DNE [correct]
Local Minimum values(s) : 48 Fixed it; works now
Saddle points(s) (x, y, f) = DNE [correct]
I think I'm putting the answer in the wrong form..?
First derivatives:
fx = y  64/x^2; fy = x  64/y^2
Second derivatives:
fxx = 128/x^3; fyy = 128/y^3; fxy = 1
Set first derivatives = 0, substitute:
y = 64/x^2; x = 64/y^2
x = 64/(64/x^2)^2 = x^4/64
Solve for x:
x(x^3  64) = 0
x = 0, x = 4
Function is undefined at x = 0.
Substituting x = 4 into fx gives y = 4.
D(x,y) = fxx(x,y)*fyy(x,y)  fxy(x,y)
D(4,4) = (2)(2)  1 = 3
Since D(4,4) > 0 and fxx(4,4) > 0, f(4,4) is a local minimum. Therefore, there are no local maximum or saddle points.
WebAssign is not accepting (4,4) as an answer. I have tried (4,4,48) [for (x,y,f) format?], f(4,4), f(4,4)=48. Is there a standard form for answering that I missed?
EDIT: 48 worked. Guess value(s) should have given it away, since it wasn't asking for a point.
I'll probably be looking for more math help next semester and post a few questions.

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